# Thread: Help with nonlinear equation - reduction of order

1. ## Help with nonlinear equation - reduction of order

Using one of the methods of reduction of order, find the general solution of the following nonlinear equation:
(y^3)y''=1

So far, using a substitution of p=y', I have managed to do:
(y^3)p(dp/dy)=1
p dp = 1/(y^3) dy
(p^2)/2 = -1/(2y^2) + c
p^2 = -1/(y^2) + c

Can someone please confirm if my working so far is correct or incorrect and where to go from here.

Thanks

2. ## Re: Help with nonlinear equation - reduction of order

Originally Posted by dwally89
Using one of the methods of reduction of order, find the general solution of the following nonlinear equation:
(y^3)y''=1

So far, using a substitution of p=y', I have managed to do:
(y^3)p(dp/dy)=1
p dp = 1/(y^3) dy
(p^2)/2 = -1/(2y^2) + c
p^2 = -1/(y^2) + c

Can someone please confirm if my working so far is correct or incorrect and where to go from here.

Thanks
So far so good...

$p=\sqrt{c -\frac{1}{y^2} }$

$p=\frac{dy}{dx}$

$\frac{dy}{dx}=\sqrt{c -\frac{1}{y^2} }$

$dx=\frac{dy}{\sqrt{c-\frac{1}{y^2}}}$

Now take integral on both sides.

3. ## Re: Help with nonlinear equation - reduction of order

Originally Posted by Also sprach Zarathustra
So far so good...

$p=\sqrt{c -\frac{1}{y^2} }$

$p=\frac{dy}{dx}$

$\frac{dy}{dx}=\sqrt{c -\frac{1}{y^2} }$

$dx=\frac{dy}{\sqrt{c-\frac{1}{y^2}}}$

Now take integral on both sides.
I managed to get to that step, but am not sure how to take the integral of the right hand side.

Thanks

4. ## Re: Help with nonlinear equation - reduction of order

Originally Posted by dwally89
I managed to get to that step, but am not sure how to take the integral of the right hand side.

Thanks
$\int\frac{dy}{\sqrt{c-\frac{1}{y^2}}}=\int \frac{y}{\sqrt{cy^2-1}}dy$

Now, substitute $u=cy^2-1$ then $du=2cy$

Try to finish...

5. ## Re: Help with nonlinear equation - reduction of order

Originally Posted by Also sprach Zarathustra
$\int\frac{dy}{\sqrt{c-\frac{1}{y^2}}}=\int \frac{y}{\sqrt{cy^2-1}}dy$

Now, substitute $u=cy^2-1$ then $du=2cy$

Try to finish...
So:
$y=\sqrt{cx^2+\frac{1}{c}}$

which then working backwards, to check that it works gives:
$y^3=({cx^2+\frac{1}{c}})^{1.5}$
$y''=({cx^2+\frac{1}{c}})^{-1.5}$
$y^3y''=1$

Right?

6. ## Re: Help with nonlinear equation - reduction of order

Originally Posted by dwally89
So:
$y=\sqrt{cx^2+\frac{1}{c}}$

which then working backwards, to check that it works gives:
$y^3=({cx^2+\frac{1}{c}})^{1.5}$
$y''=({cx^2+\frac{1}{c}})^{-1.5}$
$y^3y''=1$

Right?

I don't think so...

Steps after taking integration on both sides:

$\frac{y\sqrt{c-\frac{1}{y^2}}}{c}+A=x+B$

Or:

$y\sqrt{c-\frac{1}{y^2}}=x+K$

Or:

$\sqrt{cy^2-1}=x+K$

Or:

$cy^2-1=(x+K)^2$

Now solve for $y$.

7. ## Re: Help with nonlinear equation - reduction of order

Originally Posted by Also sprach Zarathustra
I don't think so...

Steps after taking integration on both sides:

$\frac{y\sqrt{c-\frac{1}{y^2}}}{c}+A=x+B$

Or:

$y\sqrt{c-\frac{1}{y^2}}=x+K$

Or:

$\sqrt{cy^2-1}=x+K$

Or:

$cy^2-1=(x+K)^2$

Now solve for $y$.
I have a bad habit of forgetting to add constants when integrating...

$y=\sqrt{\frac{(x+K)^2+1}{c}}$
Is that right?

8. ## Re: Help with nonlinear equation - reduction of order

Originally Posted by dwally89
$y=\sqrt{\frac{(x+K)^2+1}{c}}$
Is that right?
$y=\sqrt{\frac{(x+K)^2+1}{c}}$ or $y=-\sqrt{\frac{(x+K)^2+1}{c}}$

Thanks